Concaves 1

[Michael13]

Thanks for the help!!

Determine the intervals on which the following function is concave up or concave down.

f(x)=3x^5-25x^4-750x^3+50x

 

[Quaid]

Determine the intervals on which the following function is concave up or concave down.

f(x)=3x^5-25x^4-750x^3+50x

Hello Michael:

There is a test for concavity that uses the second derivative. You should be able to find this information in your textbook; use the index to search for keywords.

Let's start with f''(x). What is it?

Otherwise, can you explain why you're stuck?

Ciao

 

[Michael13]

Sorry all.

Sorry all, I just wanted to throw them on here real quick. This is what I have so far:

I found f''=60x^3-300x^2-4500x
I brought it all to 0 giving me x(x^2-5x-75)
this gave me x=0 for first critical point and then x^2-5x-75 gives me x=2.5
I have it as a concave down with (-infinity,0) increasing and (0,2.5)(2.5,infinity) decreasing
Is this anywhere near correct? Thanks!

 

[fcabanski]

x=0 is one critical point. But 2.5 is not a critical point. It's not a solution for x^2 - 5x -75 =0. Solve it (find the critical points) using the quadratic formula.

If you graph the original equation you can see the approximate points where the concavity changes.

 

[Michael13]

thanks for the tip.

so i am coming up with -5/2 (-1+sqrt(13)) and 5/2 (1+sqrt(13))

am i even in the ballpark?

 

[Michael13]

another thought is that when I look at the graph I can't find any of these points as the maximum or minimum so how would i determine the concavity parameters? Any help is appreciated, thank you!

 

[Michael13]

I am going with

concave down (-infinity, -5/2(sqrt13-1)),(-5/2(sqrt13-1),0)

concave up (0,5/2(sqrt13+1)

I think this nails it

 

[mmm4444bot]

when I look at the graph I can't find any of these points as the maximum or minimum

What graph are you looking at?

Why are you looking for minimum or maximum points? Does the exercise also ask for this?

 

[mmm4444bot]

I am going with

concave down (-infinity, -5/2(sqrt13-1)),(-5/2(sqrt13-1),0)

concave up (0,5/2(sqrt13+1)

This answer is not correct.

You've got the correct values for interval endpoints, but you have not accounted for the entire domain above, and you've only stated the concavity correctly for one of your intervals listed.